If a-i-j- b-2j-k- and r-a-b-a- r-b-a-b- then - r - r- is equal to

The correct option is A 1√(11) ( i⃗+3j⃗ k⃗ ) r×a=b×a r×b= [b×a] Therefore r×b= r×a r×b+r×a=0 r×(a+b)=0 Hence r is a vector p ...

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Properties of Iota

If a⃗=i⃗+j⃗...

Question

If $\to a = \to i + \to j$ , $\to b = 2 \to j - \to k$ and $\to r \times \to a = \to b \times \to a$ , $\to r \times \to b = \to a \times \to b$ , then $\frac{\to r}{| \to r |}$ is equal to

\frac{1}{\sqrt{11}} (\to i + 3 \to j - \to k)

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\frac{1}{\sqrt{11}} (\to i - 3 \to j + \to k)

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\frac{1}{\sqrt{11}} (\to i - \to j + \to k)

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None of these

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Similar questions

\to A = 2 \to i + \to k, \to B = \to i + \to j + \to k,

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\to R . \to A = 0

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(\to a \times \to b) \times \to c

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